Problem: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -2x - 2}\enspace$ and passes through the point ${(6, 6)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Explanation: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-2}$ , and its negative reciprocal is ${\dfrac{1}{2}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = \dfrac{1}{2}x + b}\enspace$ We can plug our point, $(6, 6)$ , into this equation to solve for ${b}$ , the y-intercept. $6 = {\dfrac{1}{2}}(6) + {b}$ $6 = 3 + {b}$ $6 - 3 = {b} = 3$ The equation of the perpendicular line is $\enspace {y = \dfrac{1}{2}x + 3}\enspace$. ${m = \dfrac{1}{2}, \enspace b = 3}$